Question

Why one in every 3 consecutive integers is divisible by 3

Answer 1

Let $n,n+1,n+2$ denote three consecutive integers.

When dividing an integer by 3, we can get three possible remainders: 0,1 or 2.

Therefore $n=3k \vee n=3k+1 \vee n=3k+2$ where $k\in\mathbb{Z}$.

If $n=3k$, then $n$ is divisible by 3.

If $n=3k+1$, then $n+2=3k+1+2=3k+3=3(k+1)$. Therefore $n+2$ is divisible by 3.

If $n=3k+2$, then $n+1=3k+2+1=3k+3=3(k+1)$. Therefore $n+1$ is divisible by 3.

As we can see, in each case, one of the three consecutive numbers is divisible by 3.